Re: Isomorphism

um, no. you need to prove that m = n. one way to do that, is to DEFINE yi = f(xi). since f is bijective, this shows m = n, but it does not yet show that the {f(xi)} are actually a basis. perhaps they are linearly dependent...but this can't happen because...why? perhaps they don't span all of Y....but this can't happen...why?

and again, just saying let {x1,x2,...,xn} be a basis for X, and let {y1,y2,..,yn} be a basis for Y, does not fix the problem. what is the isomorphism f for any particular (arbitrary) vector x in X?

until you say what it is, how can you even begin to assert it is a linear map (vector space homomorphism)?

Re: Isomorphism

Originally Posted by Deveno

um, no. you need to prove that m = n. one way to do that, is to DEFINE yi = f(xi). since f is bijective, this shows m = n, but it does not yet show that the {f(xi)} are actually a basis. perhaps they are linearly dependent...but this can't happen because...why? perhaps they don't span all of Y....but this can't happen...why?

and again, just saying let {x1,x2,...,xn} be a basis for X, and let {y1,y2,..,yn} be a basis for Y, does not fix the problem. what is the isomorphism f for any particular (arbitrary) vector x in X?

until you say what it is, how can you even begin to assert it is a linear map (vector space homomorphism)?